2
$\begingroup$

Background: The solution space of original problem (which requires a fine enough mesh to resolve the microstructure) can be split into a macroscale solution space and microscale solution space. This can be based on variational Multiscale Method (VMM) proposed by Hughes in 1998. By doing this, we can derive a multiscale formulation for the original heterogeneous problem. The final goal of most (multiscale) homogenisation based methods is to gain a smooth solution at the macroscale, but it also contains some information from the sub-scale. We can consider such a original problem a simple elasticity or Poisson's equation, which is bilinear form.

When I was reading a mechanics paper on two scale adaptive FE analysis for application in multicsale setting, I came across a sentence that the continuous space (i.e. the macroscale space) is the space of all fucnctions inside each RVE, that can reproduce the trace of functions in original solution space on the boundary of RVE's. So my question is what trace of functions means in PDEs. To try solving this myself, I did some online Google search and came out a Wiki link explaining trace operator trace operator, where an application into Poisson's equation is discussed.

Due to incomplete knowledge on functional analysis and measure theory, I cannot quite understand what the Wiki page is trying to describe? I attached a snapshot with my doubts highlighted in red. How to understand them? Also, you can elaborate exposition on trace of functions on boundary.

https://en.wikipedia.org/wiki/Trace_operator

$\endgroup$
6
$\begingroup$

I think the quote from Wikipedia is misleading (and I should put that on my list of things to fix :-) ). You can think of the trace of a function $u: \Omega \rightarrow {\mathbb R}$ that lives on a domain $\Omega$ as that function $g: \partial\Omega \rightarrow {\mathbb R}$ that lives on the boundary of $\Omega$ and so that $u(\mathbf x)=g(\mathbf x)$ for all points $\mathbf x$ on the boundary $\partial \Omega$. In other words, it is the restriction of $u$ to the boundary.

Things are a bit more complicated than that because $u$ may be discontinuous at individual locations, and then it's not quite obvious how one would define $g$, but fundamentally the description above is what you should think of when you think of the trace.

$\endgroup$
  • 1
    $\begingroup$ FYI, I've "fixed" the intuitive description on wikipedia. $\endgroup$ – Wolfgang Bangerth Oct 10 '17 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.